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a(n) is the least k such that A090701(k) = n.
3

%I #24 Jul 04 2023 14:00:10

%S 1,2,6,8,11,14,18,20,24,26,30,32,36,38,42,44,48,50,54,56,60,62,66,68,

%T 72,74,78,80,84,86,90,92,96,98,102,104,108,110,114,116,120,122,126,

%U 128,132,134,138,140,144,146,150,152,156,158,162,164,168,170,174

%N a(n) is the least k such that A090701(k) = n.

%H Robert Israel, <a href="/A298474/b298474.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1, 1, -1).

%F a(n) = floor(log_2(A298476(n))) + 1.

%F From _Robert Israel_, Jan 24 2018: (Start)

%F If n is even, a(n) = 3*n-4.

%F If n <> 1 or 5 is odd, a(n) = 3*n-3.

%F G.f.: x*(1+x+3*x^2+x^3-x^4+x^5+x^6-x^7)/((1-x)*(1-x^2)). (End)

%e The lexicographically earliest strings of length a(n) with a minimum palindromic partition into n parts:

%e n | a(n) | string | partition

%e --+------+----------------+---------------------------

%e 1 | 1 | 0 | (0)

%e 2 | 2 | 01 | (0)(1)

%e 3 | 6 | 001011 | (0)(010)(11)

%e 4 | 8 | 00101100 | (00)(101)(1)(00)

%e 5 | 11 | 00101100101 | (00)(101)(1001)(0)(1)

%e 6 | 14 | 00101110001011 | (00)(101)(11)(00)(010)(11)

%p f:= n -> 3*n-4+(n mod 2):

%p f(1):= 1: f(5):= 11:

%p map(f, [$1..100]); # _Robert Israel_, Jan 24 2018

%t With[{s = Array[Boole[# == 11] + Floor[#/6] + Floor[(# + 4)/6] + 1 &, 2^8]}, Array[FirstPosition[s, #][[1]] &, Max@ Take[#, LengthWhile[Differences@ #, # == 1 &]] &@ Union@ s]] (* _Michael De Vlieger_, Jan 23 2018 *)

%Y Cf. A090701, A298476.

%K nonn

%O 1,2

%A _Peter Kagey_, Jan 19 2018